JEE Main & Advanced AIEEE Solved Paper-2004

  • question_answer
    The mean and the variance of a binomial distribution are 4 and 2, respectively. Then, the probability of 2 successes is

    A) \[\frac{37}{256}\]            

    B)        \[\frac{219}{256}\]                         

    C) \[\frac{128}{256}\]                         

    D)        \[\frac{28}{256}\]

    Correct Answer: D

    Solution :

    Given that, mean\[=4\Rightarrow np=4\] and variance = 2 \[\Rightarrow \]               \[npq=2\] \[\Rightarrow \]               \[4q=2\] \[\Rightarrow \]               \[q=\frac{1}{2}\] \[\therefore \]  \[p=1-q=1-\frac{1}{2}=\frac{1}{2}\] Also, \[n=8\] Probability of 2 successes\[=P(X=2){{=}^{8}}{{C}_{2}}{{p}^{2}}{{q}^{6}}\] \[=\frac{8!}{2!6!}\times {{\left( \frac{1}{2} \right)}^{2}}\times {{\left( \frac{1}{2} \right)}^{6}}=28\times \frac{1}{{{2}^{8}}}=\frac{28}{256}\]

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